Laplace transform of Bessel's equation: $xy'' + y' + xy = 0$ page 178, "differential equations demystified", 2004:
Use the laplace transform to analyze Bessel's equation:
$$xy'' + y' + xy = 0$$
$$y(0)=1$$
We know that:
$$ L[xy] = -\frac{d}{ds}Y(s)$$
$$ L[xy'']=-\frac{d}{ds}\left(s^{2}Y(s) - s y(0)\right)$$
$$ L[y'] = s Y(s)-y(0)$$
Ok. no problem.  so far.
$$ L[xy''] + L[y'] + L[xy] = L[0]$$
$$ -\left(\frac{d}{ds}s^{2}Y(s) - s y(0)\right) + \left(s Y(s)-y(0)\right)+-\left(\frac{d}{ds}Y(s)\right) = 0$$
$$ -\left(\frac{d}{ds}s^{2}Y(s) - s\right) + \left(s Y(s)-1\right)+-\left(\frac{d}{ds}Y(s)\right) = 0$$
Now for the problem...  The textbook states that at this step, I should have this result instead:
$$ -\left(\frac{d}{ds}s^{2}Y(s) - s\right) + \left(s Y(s)-1\right)+\left(-1 - \frac{d}{ds}Y(s)\right) = 0$$
which is then simplified to:
$$(s^{2} + 1) \frac{dY(s)}{ds} = -sY(s)$$
My question is:  What is the textbook doing to get the extra "-1"?  or is the textbook wrong?
 A: The derivative $\frac{d}{ds}$ should be applied to the whole expression of $L[y'']$, not just the part $s^2 Y$. But I doubt that the answer from the text book as it is written now, is correct. That constant $-1$ in the last bracket doesn't seem right.
Since
$$
 L[y''] = s^2 Y - sy(0) - y'(0),
$$
we get
$$
 L[x y''] = -\frac{d}{ds}\left(s^2 Y - s y(0) - y'(0)\right) = -2s Y - s^2 \frac{dY}{ds} + 1.
$$
Note that by deriving, we don't need to know the constant $y'(0)$.
So the Laplace transform of the differential equation becomes
$$
  (-2 s Y - s^2 \frac{dY}{ds} + 1) +  (s Y - 1)  - \frac{dY}{ds} = 0,
$$
which becomes
$$
  -(s^2+1) \frac{dY}{ds} = s Y.
$$
A: $−1$ comes from the second derivative
\begin{align}
{\cal L}(xy'')+{\cal L}(y')+{\cal L}(xy)&=0 \\
 -[s^2{\cal L}(y)-sy(0)-y'(0)]'+[s{\cal L}(y)-y(0)]-[{\cal L}(y)]'&=0 \\
 -2s{\cal L}(y)-s^2{\cal L}'(y)+1+s{\cal L}(y)-1-{\cal L}'(y)&=0 \\
 (-s^2-1){\cal L}'(y)-s{\cal L}(y)&=0 \\
 \frac{{\cal L}'(y)}{{\cal L}(y)}&=-\frac{s}{s^2+1} \\
\end{align}
A: $$x y'' + y'+ xy = 0$$
$$L[xy'']+L[y']+L[xy]=L[0]$$
$$-\frac{d}{ds}(s^{2}Y(s)-sy(0))+(Y(s)s-y(0))-\frac{d}{ds}Y(s)=0$$
$$-(2sY(s)+s^{2}y'(s)-y(0))+(Y(s)s-y(0))-Y'(s)=0$$
$$-2sY(s)-s^{2}y'(s)+y(0)+Y(s)s-y(0)-Y'(s)=0$$
$$-2sY(s)-s^{2}y'(s)+Y(s)s-Y'(s)=0$$
$$Y(s)(-2s+s)+Y'(s)(-s^{2}-1)=0$$
$$-Y(s)s - Y'(s)(s^{2}+1)=0$$
$$Y'(s)(s^{2}+1)=-Y(s)s$$
Answer:
$$(s^{2}+1)\frac{d}{ds}Y(s)=-Y(s)s$$
A new DE reduced in order by one... now separation of variables + bionomial series expansion works to finish problem...
part 2
DE separation leads to result:
$$ Y(s) = k(s^{2}+1)^{-1/2} $$
part 3
Apply Binomial Series expansion to result of part 2
$$ {(1+x)}^\alpha=\sum_{k=0}^{\infty}{\left(\begin{matrix}\alpha\\k\\\end{matrix}\right)x^k} $$
$$ \left(\begin{matrix}\alpha\\k\\\end{matrix}\right)=\frac{1+\alpha(\alpha-1)(\alpha-2)\ \cdots\ (\alpha-k+1)}{k!}  $$
$$ {(1+x)}^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\cdots $$
part 4
inverse laplace transform of results from part 3
